How to Round to the nearest whole number in C#
C#RoundingC# Problem Overview
How can I round values to nearest integer?
For example:
1.1 => 1
1.5 => 2
1.9 => 2
"Math.Ceiling()" is not helping me. Any ideas?
C# Solutions
Solution 1 - C#
See the official documentation for more. For example:
Basically you give the Math.Round
method three parameters.
- The value you want to round.
- The number of decimals you want to keep after the value.
- An optional parameter you can invoke to use AwayFromZero rounding. (ignored unless rounding is ambiguous, e.g. 1.5)
Sample code:
var roundedA = Math.Round(1.1, 0); // Output: 1
var roundedB = Math.Round(1.5, 0, MidpointRounding.AwayFromZero); // Output: 2
var roundedC = Math.Round(1.9, 0); // Output: 2
var roundedD = Math.Round(2.5, 0); // Output: 2
var roundedE = Math.Round(2.5, 0, MidpointRounding.AwayFromZero); // Output: 3
var roundedF = Math.Round(3.49, 0, MidpointRounding.AwayFromZero); // Output: 3
You need MidpointRounding.AwayFromZero
if you want a .5 value to be rounded up. Unfortunately this isn't the default behavior for Math.Round()
. If using MidpointRounding.ToEven
(the default) the value is rounded to the nearest even number (1.5
is rounded to 2
, but 2.5
is also rounded to 2
).
Solution 2 - C#
Math.Ceiling
always rounds up (towards the ceiling)
Math.Floor
always rounds down (towards to floor)
what you are after is simply
Math.Round
which rounds as per this post
Solution 3 - C#
You need Math.Round
, not Math.Ceiling
. Ceiling
always "rounds" up, while Round
rounds up or down depending on the value after the decimal point.
Solution 4 - C#
there's this manual, and kinda cute way too:
double d1 = 1.1;
double d2 = 1.5;
double d3 = 1.9;
int i1 = (int)(d1 + 0.5);
int i2 = (int)(d2 + 0.5);
int i3 = (int)(d3 + 0.5);
simply add 0.5 to any number, and cast it to int (or floor it) and it will be mathematically correctly rounded :D
Solution 5 - C#
You can use Math.Round as others have suggested (recommended), or you could add 0.5 and cast to an int (which will drop the decimal part).
double value = 1.1;
int roundedValue = (int)(value + 0.5); // equals 1
double value2 = 1.5;
int roundedValue2 = (int)(value2 + 0.5); // equals 2
Solution 6 - C#
Just a reminder. Beware for double.
Math.Round(0.3 / 0.2 ) result in 1, because in double 0.3 / 0.2 = 1.49999999
Math.Round( 1.5 ) = 2
Solution 7 - C#
You have the Math.Round function that does exactly what you want.
Math.Round(1.1) results with 1
Math.Round(1.8) will result with 2.... and so one.
Solution 8 - C#
this will round up to the nearest 5 or not change if it already is divisible by 5
public static double R(double x)
{
// markup to nearest 5
return (((int)(x / 5)) * 5) + ((x % 5) > 0 ? 5 : 0);
}
Solution 9 - C#
I was looking for this, but my example was to take a number, such as 4.2769 and drop it in a span as just 4.3. Not exactly the same, but if this helps:
Model.Statistics.AverageReview <= it's just a double from the model
Then:
@Model.Statistics.AverageReview.ToString("n1") <=gives me 4.3
@Model.Statistics.AverageReview.ToString("n2") <=gives me 4.28
etc...
Solution 10 - C#
Use Math.Round
:
double roundedValue = Math.Round(value, 0)
Solution 11 - C#
var roundedVal = Math.Round(2.5, 0);
It will give result:
var roundedVal = 3
Solution 12 - C#
Using Math.Round(number)
rounds to the nearest whole number.
Solution 13 - C#
If your working with integers rather than floating point numbers, here is the way.
#define ROUNDED_FRACTION(numr,denr) ((numr/denr)+(((numr%denr)<(denr/2))?0:1))
Here both "numr" and "denr" are unsigned integers.
Solution 14 - C#
Write your own round method. Something like,
function round(x) rx = Math.ceil(x) if (rx - x <= .000001) return int(rx) else return int(x) end
Solution 15 - C#
decimal RoundTotal = Total - (int)Total;
if ((double)RoundTotal <= .50)
Total = (int)Total;
else
Total = (int)Total + 1;
lblTotal.Text = Total.ToString();